The Real Line — Worked Problems

What this book adds

The free chapters on this topic teach the ideas and the traps. This is the companion problem set: eleven questions at the length and difficulty CSIR NET actually sets, each solved completely rather than sketched — followed by a compressed reference sheet and six problems for you to check yourself against, answers only.

Work them in order. Several later solutions reuse a result from an earlier one, the way an exam paper itself rewards having done the easy part first.

Quick reference

Supremum. $s = \sup A$ iff $s$ is an upper bound and, for every $\varepsilon > 0$, some $a \in A$ has $a > s - \varepsilon$. Not necessarily attained.

Archimedean property. For every $x \in \mathbb{R}$ there is $n \in \mathbb{N}$ with $n > x$.

The implication chain. convergent $\Rightarrow$ Cauchy $\Rightarrow$ bounded $\Rightarrow$ has a convergent subsequence. Only Cauchy $\Rightarrow$ convergent needs completeness; none of the arrows reverse without it.

Monotone convergence. Increasing and bounded above $\Rightarrow$ converges, to the supremum. Prove monotone, prove bounded, then solve $L = f(L)$ — never before.

limsup / liminf. $\limsup a_n = \lim_{n} \sup_{k \ge n} a_k$, the largest subsequential limit. $\liminf a_n \le \limsup a_n$ always; equality (and finite) iff the sequence converges. Subadditive only: $\limsup(a_n+b_n) \le \limsup a_n + \limsup b_n$, with equality restored the moment one of the two sequences converges.

Choosing a series test, in order: $a_n \not\to 0 \Rightarrow$ diverges, free. Shape $1/n^p$ or geometric $\Rightarrow$ comparison ($\sum 1/n^p$ converges iff $p>1$). Factorials or $n^n$ $\Rightarrow$ ratio. $n$-th powers $\Rightarrow$ root (strictly stronger than ratio). Ratio $\to 1$ $\Rightarrow$ Raabe: $R = \lim n(a_n/a_{n+1} - 1)$, converges for $R>1$, diverges for $R<1$. Logs, terms decreasing $\Rightarrow$ Cauchy condensation: $\sum a_n$ and $\sum 2^k a_{2^k}$ share a fate. Alternating $\Rightarrow$ Leibniz, then check $\sum |a_n|$ separately for absolute convergence.

Worked problems

1. A nested family of open intervals whose intersection is closed

Problem. For $n \in \mathbb{N}$, let $A_n = \left(-\dfrac{1}{n},\ 1 + \dfrac{1}{n}\right)$. Find $\bigcup_{n=1}^{\infty} A_n$ and $\bigcap_{n=1}^{\infty} A_n$.

Solution.

Strategy: show the family is nested and decreasing, so the union is just the largest set; then pin down the intersection by checking $[0,1]$ survives and nothing outside it does.

As $n$ grows, $-1/n$ increases toward $0$ while $1+1/n$ decreases toward $1$, so each interval sits entirely inside the one before it.

Key fact. The family is nested and decreasing: $A_1 \supseteq A_2 \supseteq A_3 \supseteq \cdots$.

For a decreasing nested family the union is simply the first, largest set — nothing outside $A_1$ is ever reached at any stage.

Union.

$$\bigcup_{n=1}^{\infty} A_n = A_1 = (-1, 2).$$

The intersection needs more care. First, $[0,1] \subseteq \bigcap_{n=1}^{\infty} A_n$: for every $n$, $0 > -1/n$ and $1 < 1+1/n$, so both endpoints of $[0,1]$ lie in every $A_n$, and so does everything between them.

Case $x<0$: write $x = -\varepsilon$ with $\varepsilon > 0$.

Tool invoked: the Archimedean property. There is $n$ with $n > 1/\varepsilon$, i.e. $1/n < \varepsilon$, so $-1/n > -\varepsilon = x$ and $x \notin A_n$.

Case $x>1$: the symmetric argument, using the upper endpoint $1+1/n$ in place of $-1/n$, excludes it the same way.

So no point outside $[0,1]$ survives every $A_n$.

Final answer.

$$\bigcup_{n=1}^{\infty} A_n = (-1, 2), \qquad \bigcap_{n=1}^{\infty} A_n = [0, 1].$$

In words: the union is the open interval from $-1$ to $2$, and the intersection is the closed interval from $0$ to $1$.

Every $A_n$ is open, and their intersection is closed. This is not a contradiction — only a finite intersection of open sets is guaranteed to be open; an infinite one need not be, and this is the standard example examiners reach for when a statement quietly drops the word "finite."

2. A supremum that is attained and an infimum that is not, in the same set

Problem. Let $A = \{(-1)^n + \tfrac{1}{n} : n \in \mathbb{N}\}$. Find $\sup A$ and $\inf A$, and say whether each is attained.

Solution.

Strategy: split the sequence into its even and odd tracks, find the extreme value each track reaches, then compare the two tracks to see which extreme is attained and which is only approached.

Split on parity: the factor $(-1)^n$ makes the even-indexed and odd-indexed terms behave completely differently, so treat them as two separate tracks.

For even $n$: $(-1)^n + 1/n = 1 + 1/n$, largest at the smallest even $n$, namely $n=2$, and it decreases toward $1$ (never reaching it) as $n$ grows further.

Even track's peak.

$$1 + \frac{1}{2} = \frac{3}{2}.$$

For odd $n$: $(-1)^n + 1/n = -1 + 1/n$, largest at $n=1$, and it decreases toward $-1$ (never reaching it) as $n$ grows further.

Odd track's peak.

$$-1 + \frac{1}{1} = 0.$$

Comparing the two tracks, the overall largest value is $3/2$ from the even track, attained at $n=2$; the overall smallest values come from the odd track, which descends toward $-1$ without ever reaching it, since equality would require $1/n=0$.

Final answer.

$$\sup A = \frac{3}{2}, \qquad \inf A = -1.$$

In words: the supremum $3/2$ is attained, at $n=2$; the infimum $-1$ is the greatest lower bound but is never actually reached.

There is no rule that a set's supremum and infimum are either both attained or both not — each is decided separately, by its own track, and this set is exactly the kind of example built to make that point.

3. Density of $\mathbb{Q}$ in $\mathbb{R}$, from the Archimedean property

Problem. Prove that for any reals $a < b$, there is a rational number $r$ with $a < r < b$.

Solution.

Strategy: use the Archimedean property twice — once to shrink a reciprocal below the gap $b-a$, and once to pick the least integer just above $na$ — then show the resulting candidate $r=m/n$ is forced strictly between $a$ and $b$.

Since $b - a > 0$, the Archimedean property gives an $n$ large enough that its reciprocal undercuts the gap.

Archimedean bound. $n \in \mathbb{N}$ with $n > \dfrac{1}{b-a}$, i.e. $\dfrac{1}{n} < b - a$.

Let $m$ be the smallest integer with $m > na$; such an $m$ exists because the Archimedean property gives some integer above $na$, and a non-empty set of integers bounded below has a least element. Minimality of $m$ then gives a second bound.

Bounds on the integer. $m > na$, and by minimality $m - 1 \le na$, i.e. $m \le na + 1$.

Now put $r = m/n$; the two bounds above pin down where $r$ sits relative to $a$ and $b$.

Locating the candidate. From $m > na$: $r = m/n > a$. From $m \le na+1$: $r = m/n \le a + 1/n < a + (b-a) = b$.

So $a < r < b$ with $r = m/n \in \mathbb{Q}$.

Final answer.

$$a < r < b, \qquad r = \frac{m}{n} \in \mathbb{Q}.$$

In words: the rational $r=m/n$, built from the two Archimedean choices, lies strictly between $a$ and $b$ — so the rationals are dense in $\mathbb{R}$.

This construction works no matter how short the interval $(a,b)$ is — the reals fill every gap densely enough that the Archimedean property alone finds a rational inside it.

4. A recursively defined sequence, proved monotone and bounded, limit found last

Problem. Define $a_1 = 2$ and $a_{n+1} = \dfrac{1}{2}\left(a_n + \dfrac{2}{a_n}\right)$ for $n \ge 1$. Show $(a_n)$ converges, and find its limit.

Solution.

Strategy: bound the sequence below by $\sqrt2$ via AM–GM, show it is decreasing, invoke the monotone convergence theorem to secure a limit, and only then solve for that limit's value.

AM–GM applied to any $x>0$ gives $\dfrac{1}{2}(x + 2/x) \ge \sqrt{x \cdot 2/x} = \sqrt2$, so $a_{n+1} \ge \sqrt2$ for every $n \ge 1$, and $a_1 = 2 \ge \sqrt2$ as well.

Lower bound.

$$a_n \ge \sqrt2, \qquad n \ge 1.$$

Next, compute the step-to-step difference to check monotonicity.

Monotonicity.

$$a_{n+1} - a_n = \frac{1}{2}\left(\frac{2}{a_n} - a_n\right) = \frac{2 - a_n^2}{2a_n}, \qquad a_n \ge \sqrt2 \implies a_n^2 \ge 2 \implies 2-a_n^2 \le 0 \implies a_{n+1} \le a_n.$$

So $(a_n)$ is decreasing and bounded below by $\sqrt2$.

Theorem invoked. A decreasing sequence bounded below converges, by the monotone convergence theorem — call the limit $L$.

Only now, with existence established, take the limit of both sides of the recursion.

Solving for the limit.

$$L = \frac{1}{2}\left(L + \frac{2}{L}\right) \implies 2L^2 = L^2 + 2 \implies L^2 = 2.$$

Since every $a_n \ge \sqrt2 > 0$, $L \ge \sqrt2 > 0$.

Final answer.

$$L = \sqrt2.$$

In words: the sequence converges, and its limit is $\sqrt2$.

This is Newton's method for $\sqrt2$ starting from $a_1=2$; it is also why the sequence converges so fast — $a_4$ already agrees with $\sqrt2$ to five decimal places.

5. Proving a sequence is Cauchy directly, without assuming a limit exists

Problem. Let $s_n = \displaystyle\sum_{k=1}^{n} \frac{1}{k^2}$. Prove $(s_n)$ is Cauchy.

Solution.

Strategy: bound the tail $|s_m-s_n|$ above by a telescoping sum that collapses to $1/n$, then use the Archimedean property to make that bound smaller than any given $\varepsilon$.

Take $m > n$; bound each term $1/k^2$ by $1/(k(k-1))$, valid for every $k \ge 2$, and let the resulting sum telescope.

Telescoping bound.

$$|s_m - s_n| = \sum_{k=n+1}^{m} \frac{1}{k^2} < \sum_{k=n+1}^{m} \frac{1}{k(k-1)} = \sum_{k=n+1}^{m} \left(\frac{1}{k-1} - \frac{1}{k}\right) = \frac{1}{n} - \frac{1}{m} < \frac{1}{n}.$$

Tool invoked: the Archimedean property. Given $\varepsilon > 0$, there is $N$ with $1/N < \varepsilon$.

Final answer.

$$m > n \ge N \implies |s_m - s_n| < \frac{1}{n} \le \frac{1}{N} < \varepsilon.$$

In words: past the index $N$, any two partial sums are closer together than $\varepsilon$, however small $\varepsilon$ is chosen — exactly the Cauchy condition, so $(s_n)$ is Cauchy.

This is where completeness earns its keep — Cauchy in $\mathbb{R}$ means convergent, with no need to name the limit. (It happens to be $\pi^2/6$, which is a separate, much harder fact.)

The technique — bound the tail by a telescoping sum — is the standard route to proving a sequence of partial sums is Cauchy directly, and it works whenever the terms are eventually dominated by $1/(k(k-1))$ or $1/(k(k+1))$.

6. One sequence, four questions and both limits

Problem. Let $a_n = \dfrac{(-1)^n n}{n+1}$. Determine whether $(a_n)$ is (i) bounded, (ii) Cauchy, (iii) convergent, (iv) has a convergent subsequence — and compute $\limsup a_n$ and $\liminf a_n$.

Solution.

Strategy: split on parity to find the sequence's two subsequential limits, then read every one of the four yes/no questions off those two tracks — the limsup/liminf question is not a fifth computation, it is the same two numbers already found.

Split on parity again, since $(-1)^n$ is exactly the kind of factor that makes this the right move.

Tracks. Even $n$: $a_n = \dfrac{n}{n+1} \to 1$. Odd $n$: $a_n = -\dfrac{n}{n+1} \to -1$.

(i) Bounded. $|a_n| = n/(n+1) < 1$ for every $n$, so yes.

(iv) Convergent subsequence. Yes — in fact two: the even-indexed terms converge to $1$ and the odd-indexed terms converge to $-1$.

(iii) Convergent. No. A sequence with two distinct subsequential limits cannot converge — if it did, every subsequence would converge to that same limit.

(ii) Cauchy. No, for the same reason: in $\mathbb{R}$, Cauchy and convergent coincide, and this sequence is not convergent.

limsup and liminf. The subsequential limits are exactly $1$ and $-1$, and there are no others, so $\limsup a_n = 1$ and $\liminf a_n = -1$ — consistent with $\liminf < \limsup$, which must hold given the sequence does not converge.

Final answer. (i) bounded — yes; (ii) Cauchy — no; (iii) convergent — no; (iv) has a convergent subsequence — yes; $\limsup a_n = 1$, $\liminf a_n = -1$. In words: the sequence is bounded and has convergent subsequences (to $1$ along even indices, $-1$ along odd indices), but since those two limits differ it is neither Cauchy nor convergent.

This is the shape almost every "decide everything about this sequence" question takes: find the tracks, read every answer off them, and the limsup/liminf question is not a fifth calculation but the same two numbers already found in part (iv).

7. limsup and liminf from residue-class tracks

Problem. Compute $\limsup a_n$ and $\liminf a_n$ for $a_n = \cos\!\left(\dfrac{n\pi}{3}\right) + \dfrac{1}{n}$.

Solution.

Strategy: reduce the periodic part $\cos(n\pi/3)$ to its six residue classes mod $6$, let the vanishing $1/n$ term settle each track to its limit, and read the extreme two off as limsup and liminf.

$\cos(n\pi/3)$ depends only on $n \bmod 6$, cycling through six values.

Residue-class values.

$$n \equiv 0,1,2,3,4,5 \pmod 6 \ \longmapsto\ \cos\frac{n\pi}{3} = 1,\ \tfrac12,\ -\tfrac12,\ -1,\ -\tfrac12,\ \tfrac12.$$

The added $1/n \to 0$ shifts every track down toward its limiting value without changing which values are largest and smallest.

Extreme tracks. Along $n \equiv 0 \pmod 6$: $a_n = 1 + 1/n \to 1$. Along $n \equiv 3 \pmod 6$: $a_n = -1 + 1/n \to -1$.

No other residue class reaches outside $[-1/2, 1/2]$ in the limit, so these two are the extreme subsequential limits.

Final answer.

$$\limsup a_n = 1, \qquad \liminf a_n = -1.$$

In words: the sequence's largest limiting value is $1$, reached along $n\equiv0\pmod6$, and its smallest is $-1$, reached along $n\equiv3\pmod6$.

This is the general method for any sequence built from a periodic piece plus a term tending to $0$: reduce to residue classes mod the period, take each track to its limit, and read off the largest and smallest.

8. When does subadditivity become equality?

Problem. Prove: if $a_n \to L$ (a genuine limit, not just bounded) and $(b_n)$ is any bounded sequence, then $\limsup(a_n + b_n) = L + \limsup b_n$.

Solution.

Strategy: sandwich $\limsup(a_n+b_n)$ between $L+M-\varepsilon$ and $L+M+\varepsilon$ for every $\varepsilon>0$, using that $a_n$ is eventually within $\varepsilon/2$ of its genuine limit $L$.

Let $M = \limsup b_n$ and fix $\varepsilon > 0$; since $a_n \to L$, there is $N$ with $|a_n - L| < \varepsilon/2$ for all $n \ge N$.

Upper bound. For $n \ge N$:

$$\sup_{k \ge n}(a_k+b_k) \le \sup_{k\ge n}\big((L+\varepsilon/2)+b_k\big) = L + \varepsilon/2 + \sup_{k \ge n} b_k.$$

Letting $n \to \infty$ gives $\limsup(a_n+b_n) \le L + \varepsilon/2 + M$.

Lower bound. Symmetrically, $a_k > L - \varepsilon/2$ for $k \ge N$ gives, for $n \ge N$:

$$\sup_{k\ge n}(a_k+b_k) \ge L - \varepsilon/2 + \sup_{k\ge n} b_k,$$

so $\limsup(a_n+b_n) \ge L - \varepsilon/2 + M$.

Both bounds hold for every $\varepsilon > 0$, so $\limsup(a_n+b_n) = L + M$.

Final answer.

$$\limsup(a_n+b_n) = L + \limsup b_n.$$

In words: once one sequence has a genuine limit $L$, the usual subadditive inequality $\limsup(a_n+b_n)\le\limsup a_n+\limsup b_n$ tightens to equality.

Why this does not contradict $\limsup(a_n{+}b_n) \le \limsup a_n + \limsup b_n$ sometimes being strict, as in $(-1)^n$ against $(-1)^{n+1}$: that example has neither sequence converging.

The moment one side is forced to a genuine limit, it can no longer be out of phase with the other, and the inequality tightens to equality. Strictness is entirely a phase effect between two oscillating sequences — remove the oscillation from one side and it is gone.

9. Raabe's test, on a series the ratio test cannot settle

Problem. Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{(2n)!}{4^n (n!)^2}$ converges.

Solution.

Strategy: show the ordinary ratio test is inconclusive here, then apply Raabe's test to the same ratio to get a definite verdict.

Let $a_n = \dfrac{(2n)!}{4^n(n!)^2}$; compute the ratio.

Ratio test — inconclusive.

$$\frac{a_{n+1}}{a_n} = \frac{(2n+2)!}{4^{n+1}((n+1)!)^2} \cdot \frac{4^n(n!)^2}{(2n)!} = \frac{(2n+1)(2n+2)}{4(n+1)^2} = \frac{2n+1}{2n+2} \longrightarrow 1.$$

The ratio tending to $1$ is exactly the signal to reach for Raabe's test; take the reciprocal ratio.

Raabe's test.

$$\frac{a_n}{a_{n+1}} = \frac{2n+2}{2n+1} = 1 + \frac{1}{2n+1}, \qquad R = \lim_{n\to\infty} n\left(\frac{a_n}{a_{n+1}} - 1\right) = \lim_{n\to\infty} \frac{n}{2n+1} = \frac12.$$

Since $R = 1/2 < 1$, the series diverges.

Final answer.

$$R = \frac12 < 1.$$

In words: Raabe's index comes out below $1$, so the series $\sum (2n)!/(4^n(n!)^2)$ diverges.

This matches the asymptotic $a_n \sim 1/\sqrt{\pi n}$ from Stirling's formula — a $p=1/2$ comparison, well under the $p>1$ threshold — but Raabe reaches the same conclusion without needing Stirling at all, which is the actual point of having the test.

10. Cauchy condensation, worked in general rather than by example

Problem. For $n \ge 2$ and $p > 0$, decide for which $p$ the series $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n(\log n)^p}$ converges.

Solution.

Strategy: confirm the terms are positive and decreasing so Cauchy condensation applies, compute the condensed series explicitly, and read convergence off the resulting $p$-series in $k$.

The terms $a_n = 1/(n(\log n)^p)$ are positive and decreasing, since both $n$ and $\log n$ increase.

Theorem invoked: Cauchy condensation. Since $(a_n)$ is positive and decreasing, $\sum a_n$ converges iff $\sum 2^k a_{2^k}$ converges.

Compute the condensed term directly.

Condensed term.

$$2^k a_{2^k} = \frac{2^k}{2^k(\log 2^k)^p} = \frac{1}{(k \log 2)^p} = \frac{1}{(\log 2)^p}\cdot\frac{1}{k^p}.$$

Dropping the constant $1/(\log2)^p$, this is a $p$-series in $k$, which converges iff $p>1$.

Final answer.

$$\sum_{n=2}^{\infty} \frac{1}{n(\log n)^p} \text{ converges} \iff p > 1.$$

In words: the series converges exactly when the log-power exponent exceeds $1$, and diverges otherwise.

In particular $\sum 1/(n\log n)$ ($p=1$) diverges and $\sum 1/(n\log^2 n)$ ($p=2$) converges — the specific pair the exam likes to set side by side is just this general result at two values of $p$, not two facts to memorise separately.

11. Absolute versus conditional, reusing Problem 10

Problem. Determine whether $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{n\log n}$ converges absolutely, converges conditionally, or diverges.

Solution.

Strategy: check absolute convergence first by reusing Problem 10 at $p=1$, then check convergence itself via Leibniz to land on conditional convergence.

Absolute values give $\sum 1/(n \log n)$, which is exactly Problem 10 at $p=1$.

Not absolutely convergent. $\sum 1/(n\log n)$ diverges, by Problem 10 at $p=1$.

For convergence itself, apply Leibniz to $a_n = 1/(n\log n)$: it is positive, decreasing for $n \ge 2$ (as argued in Problem 10), and $a_n \to 0$.

Leibniz applies. The alternating series $\sum (-1)^n/(n\log n)$ converges.

Convergent, but not absolutely convergent.

Final answer. The series $\sum (-1)^n/(n\log n)$ is conditionally convergent: convergent by Leibniz, but not absolutely convergent since $\sum 1/(n\log n)$ diverges. In words: it converges only because of the alternating signs, not outright.

And because it is only conditionally convergent, Riemann's rearrangement theorem applies to it — reordering its terms can be made to sum to any real number at all, or to diverge, which is not true of Problem 5's series (that one converges absolutely, so every rearrangement sums to the same value).

Practice — check yourself

Answers only — the point is to test whether the methods above transfer, not to read another eleven solutions.

1. Find $\sup B$ and $\inf B$, and say whether each is attained, for $B = \{1 - 1/n : n \in \mathbb{N}\} \cup \{2\}$.
2. Is $a_n = (-1)^n\left(1 + \dfrac{1}{n^2}\right)$ Cauchy?
3. Compute $\limsup a_n$ and $\liminf a_n$ for $a_n = (-1)^n + \dfrac{1}{n}\sin\dfrac{n\pi}{2}$.
4. Does $\displaystyle\sum_{n=1}^\infty \frac{n!}{n^n}$ converge? (Ratio test.)
5. Is $\displaystyle\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt n}$ absolutely convergent, conditionally convergent, or divergent?
6. For $c_n = \sup\{1/k : k \ge n\}$, find $\lim_n c_n$.

Work them before reading on — the answers are right below.

Answers

1. $\sup B = 2$ (attained, it is the isolated point); $\inf B = 0$ (attained, at $n=1$).
2. No — the even and odd subsequences converge to $1$ and $-1$ respectively, so the sequence does not converge, hence is not Cauchy.
3. $\limsup a_n = 1$, $\liminf a_n = -1$ (the $\frac1n\sin(n\pi/2)$ term is squeezed to $0$ and does not move either extreme).
4. Converges: $a_{n+1}/a_n = (n/(n+1))^n \to 1/e < 1$.
5. Conditionally convergent: $\sum 1/\sqrt n$ is a $p=1/2$ series and diverges, but the alternating series itself converges by Leibniz.
6. $0$ — the tail supremum of a decreasing sequence is its first term, so $c_n = 1/n \to 0$.